If int e^x cos x , dx = frac{e^x}{2}(cos x + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx$.Substitute $t = \log \left(\frac{2x+3}{3-2x}\right)$,

Vedclass · Accepted Answer

$\log f(1)$

Vedclass · Answer

(C) Let $I = \int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx$.Substitute $t = \log \left(\frac{2x+3}{3-2x}\right)$, so $\frac{2x+3}{3-2x} = e^t$.Differentiating both sides with respect to $x$: $\frac{d}{dx} \left( \frac{2x+3}{3-2x} \right) = \frac{(3-2x)(2) - (2x+3)(-2)}{(3-2x)^2} = \frac{6-4x+4x+6}{(3-2x)^2} = \frac{12}{(3-2x)^2}$.Thus, $\frac{12}{(3-2x)^2} \, dx = e^t \, dt$, which implies $\frac{1}{(3-2x)^2} \, dx = \frac{1}{12} e^t \, dt$.Substituting into the integral: $I = \int \cos(t) \cdot \frac{1}{12} e^t \, dt = \frac{1}{12} \int e^t \cos(t) \, dt$.Using the given formula $\int e^t \cos(t) \, dt = \frac{e^t}{2}(\cos t + \sin t)$, we get $I = \frac{1}{12} \cdot \frac{e^t}{2}(\cos t + \sin t) = \frac{e^t}{24}(\cos t + \sin t)$.Substituting back $t = \log \left(\frac{2x+3}{3-2x}\right)$ and $e^t = \frac{2x+3}{3-2x}$, we have $I = \frac{\frac{2x+3}{3-2x}}{24} \left[ \cos \left( \log \left( \frac{2x+3}{3-2x} \right) \right) + \sin \left( \log \left( \frac{2x+3}{3-2x} \right) \right) \right] + c$.Comparing with the given form, $f(x) = \frac{2x+3}{3-2x}$ and $g(x) = \log \left( \frac{2x+3}{3-2x} \right)$.Therefore, $g(x) = \log(f(x))$, which implies $g(1) = \log(f(1))$.

If $\int e^x \cos x \, dx = \frac{e^x}{2}(\cos x + \sin x)$ and $\int \frac{\cos \left(\log \left(\frac{2x+3}{3-2x}\right)\right)}{(3-2x)^2} \, dx = \frac{f(x)}{24}[\cos (g(x)) + \sin (g(x))] + c$, then $g(1) =$

Similar Questions

If $\int \frac{\sin x}{\sin (x-\alpha)} dx = px - q \log |\sin (x-\alpha)| + c$,then $pq =$ . . . . . . .

The value of the integral $\int_0^1 \frac{x^b - 1}{\log x} \, dx$ is

Let $I(x)=\int\frac{3dx}{(4x+6)(\sqrt{4x^{2}+8x+3})}$ and $I(0)=\frac{\sqrt{3}}{4}+20$. If $I(\frac{1}{2})=\frac{a\sqrt{2}}{b}+c$, where $a, b, c \in N$ and $gcd(a,b)=1$, then $a+b+c$ is equal to:

The value of $\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is

If $\int(x^6+x^4+x^2) \sqrt{2x^4+3x^2+6} dx = f(x) + c$,then $f(3) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module